I'am trying to integrate the following function$$\int \frac{2}{(v-1)^2}dv$$ So far I've tried $u$ substitution.Here are the steps.Let $$u = (v-1)^2$$ then$$\frac{du}{dv} = 2v-2$$ solving for $dv$ $$dv = \frac{1}{2(v-1)}du$$ This doesn't seem to me a sensible way of working out this integration.I'am stuck Please help.
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3 Answers
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2 Let $u=v-1$, $du=dv$. Then
$$\int\frac{dv}{(v-1)^2}=\int\frac{du}{u^2}.$$
I'm sure you can finish, now!
- 1$\begingroup$ It's $\frac{-2}{v-1} + C$.Yeah Bruno Rocks! $\endgroup$alok– alok2011-12-27 19:50:36 +00:00Commented Dec 27, 2011 at 19:50
- $\begingroup$ I do my best! :-) $\endgroup$Bruno Joyal– Bruno Joyal2011-12-27 19:52:50 +00:00Commented Dec 27, 2011 at 19:52
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Letting $u=v-1$ makes the integral $\int2u^{-2}du$, which I bet you can do on your own.
Always consider the possibility of the simplest substitutions (e.g. linear) before anything else.
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1 Are you familiar with the power rule for integration, $$\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\quad (n\neq 1)\quad ?$$ There is a different $u$-substitution you can make where you can apply it.
- $\begingroup$ yes i'am familiar $\endgroup$alok– alok2011-12-27 19:44:52 +00:00Commented Dec 27, 2011 at 19:44